/* Based on http://en.wikipedia.org/wiki/Quartic_function#Quick_and_memorable_solution_from_first_principles */
static int solve_depressed_quartic(const complex_t* poly, complex_t* results) {
complex_t helper_cubic[4];
complex_t helper_results[3];
complex_t quadratic_factor[3];
complex_t p, c_plus_p_sqr, d_div_p;
const complex_t e = poly[0];
const complex_t d = poly[1];
const complex_t c = poly[2];
int num_results;
if (complex_eq(d, complex_from_real(0.0))) {
int i, num_quad_results;
complex_t quadratic[3];
complex_t quadratic_results[2];
quadratic[0] = e;
quadratic[1] = c;
quadratic[2] = complex_from_real(1.0);
num_quad_results = solve_poly(2, quadratic, quadratic_results);
for (i = 0; i < num_quad_results; ++i) {
const complex_t s = complex_sqrt(quadratic_results[i]);
results[2*i] = complex_negate(s);
results[2*i + 1] = s;
}
return 2 * num_quad_results;
}
helper_cubic[0] = complex_negate(complex_mult(d, d));
helper_cubic[1] = complex_add(complex_mult(c, c), complex_mult_real(-4.0, e));
helper_cubic[2] = complex_mult_real(2.0, c);
helper_cubic[3] = complex_from_real(1.0);
if (solve_poly(3, helper_cubic, helper_results) < 1)
return 0;
p = complex_sqrt(helper_results[0]);
c_plus_p_sqr = complex_add(c, complex_mult(p, p));
d_div_p = complex_div(d, p);
quadratic_factor[0] = complex_add(c_plus_p_sqr, complex_negate(d_div_p));
quadratic_factor[1] = complex_mult_real(2.0, p);
quadratic_factor[2] = complex_from_real(2.0);
num_results = solve_poly(2, quadratic_factor, results);
quadratic_factor[0] = complex_add(c_plus_p_sqr, d_div_p);
quadratic_factor[1] = complex_negate(quadratic_factor[1]);
return num_results + solve_poly(2, quadratic_factor, results + num_results);
}
/* Based on http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method */
static int solve_depressed_cubic(const complex_t* poly, complex_t* results) {
const complex_t q = poly[0];
const complex_t p = poly[1];
complex_t t, u, cubic_root_of_unity;
int i;
if (complex_eq(p, complex_from_real(0.0))) {
results[0] = complex_pow_real(complex_negate(q), 1.0/3.0);
return 1;
}
t = complex_add(
complex_mult_real(0.25, complex_mult(q, q)),
complex_mult_real(1.0/27.0, complex_mult(p, complex_mult(p, p))));
cubic_root_of_unity.real = -0.5;
cubic_root_of_unity.imag = 0.5 * sqrt(3.0);
for (i = 0; i < 3; ++i) {
if (i == 0)
u = complex_pow_real(complex_add(complex_mult_real(-0.5, q), complex_sqrt(t)), 1.0/3.0);
else
u = complex_mult(u, cubic_root_of_unity);
results[i] = complex_add(u, complex_div(p, complex_mult_real(-3.0, u)));
}
return 3;
}
static int solve_depressed_quadratic(const complex_t* poly, complex_t* results) {
const complex_t t = complex_sqrt(complex_negate(poly[0]));
results[0] = complex_negate(t);
results[1] = t;
return 2;
}
static void initial_tetrahedron(
Triangulation *manifold,
Tetrahedron **initial_tet,
Boolean compute_corners,
Boolean centroid_at_origin)
{
VertexIndex v[4];
Complex z,
sqrt_z,
w[4];
Tetrahedron *tet;
EdgeIndex best_edge,
edge;
/*
* Set a default choice of tetrahedron and edge.
*/
*initial_tet = manifold->tet_list_begin.next;
best_edge = 0;
/*
* 2000/02/11 JRW Can we choose the initial tetrahedron in such
* a way that if we happen to have the canonical triangulation
* of a 2-bridge knot or link complement, the basepoint falls
* at a center of D2 symmetry? That is, can we find a Tetrahedron
* that looks like the "top of the tower" in the canonical
* triangulation of a 2-bridge knot or link complement?
*/
for (tet = manifold->tet_list_begin.next;
tet != &manifold->tet_list_end;
tet = tet->next)
for (edge = 0; edge < 6; edge++)
if (tet->neighbor[one_face_at_edge [edge]]
== tet->neighbor[other_face_at_edge[edge]])
{
*initial_tet = tet;
best_edge = edge;
}
if (compute_corners)
{
if (centroid_at_origin == TRUE)
{
/*
* Proposition. For any value of w, positioning the corners at
*
* corner[0] = w
* corner[1] = w^-1
* corner[2] = -w^-1
* corner[3] = -w
*
* defines a tetrahedron with its centroid at the "origin" and
* the common perpendiculars between pairs of opposite edges
* coincident with the "coordinate axes". [In the Klein model,
* the tetrahedron is inscribed in a rectangular box whose faces
* are parallel to the coordinate axes.]
*
* Proof: Use the observation that the line from a0 to a1 will
* intersect the line from b0 to b1 iff the cross ratio
*
* (b0 - a0) (b1 - a1)
* -------------------
* (b1 - a0) (b0 - a1)
*
* of the tetrahedron they span is real, and they will be
* orthogonal iff the cross ratio is -1.
*
* [-w, w] is orthogonal to [0, infinity] because
*
* (0 - -w) (infinity - w)
* ----------------------- = -1
* (infinity - -w) (0 - w)
*
* and similarly for [-w^-1, w^-1] and [0, infinity].
*
* [w^-1, w] is orthogonal to [-1, 1] because
*
* (-1 - w^-1) (1 - w)
* ------------------- = -1
* (1 - w^-1) (-1 - w)
*
* and similarly for [-w^-1, -w] and [-1, 1].
*
* [-w^-1, w] is orthogonal to [-i, i] because
*
* (-i - -w^-1) (i - w)
* -------------------- = -1
* (i - -w^-1) (-i - w)
*
* and similarly for [w^-1, -w] and [-i, i].
*
* Q.E.D.
*
*
* The tetrahedron will have the correct cross ratio z iff
*
* (w - -w^-1) (w^-1 - -w ) (w + w^-1)^2
* z = -------------------------- = --------------
* (w - -w ) (w^-1 - -w^-1) 4
*
* Solving for w in terms of z gives the four possibilities
*
* w = +- (sqrt(z) +- sqrt(z - 1))
*
* Note that sqrt(z) + sqrt(z - 1) and sqrt(z) - sqrt(z - 1) are
* inverses of one another. We can choose any of the four solutions
* to be "w", and the other three will automatically become w^-1,
* -w, and -w^-1.
*
* Comment: This position for the initial corners brings out
* nice numerical properties in the O(3,1) matrices for manifolds
* composed of regular ideal tetrahedra (cf. the proofs in the
* directory "Tilings of H^3", which aren't part of SnapPea, but
* I could give you a copy).
*/
z = (*initial_tet)->shape[filled]->cwl[ultimate][0].rect;
w[0] = complex_plus(
complex_sqrt(z),
complex_sqrt(complex_minus(z, One))
);
w[1] = complex_div(One, w[0]);
w[2] = complex_negate(w[1]);
w[3] = complex_negate(w[0]);
(*initial_tet)->corner[0] = w[0];
(*initial_tet)->corner[1] = w[1];
(*initial_tet)->corner[2] = w[2];
(*initial_tet)->corner[3] = w[3];
}
else
{
/*
* Originally this code positioned the Tetrahedron's vertices
* at {0, 1, z, infinity}. As of 2000/02/04 I modified it
* to put the vertices at {0, 1/sqrt(z), sqrt(z), infinity} instead,
* so that the basepoint (0,0,1) falls at the midpoint
* of the edge extending from 0 to infinity, and the
* tetrahedron's symmetry axis lies parallel to the x-axis.
* To convince yourself that the tetrahedron's axis of
* symmetry does indeed pass through that point, note
* that a half turn around the axis of symmetry factors
* as a reflection in the plane |z| = 1 followed by
* a reflection in the vertical plane sitting over x-axis.
*/
/*
* Order the vertices so that the tetrahedron is positively
* oriented, and the selected edge is between vertices
* v[0] and v[1].
*/
v[0] = one_vertex_at_edge[best_edge];
v[1] = other_vertex_at_edge[best_edge];
v[2] = remaining_face[v[1]][v[0]];
v[3] = remaining_face[v[0]][v[1]];
/*
* Set the coordinates of the corners.
*/
z = (*initial_tet)->shape[filled]->cwl[ultimate][edge3[best_edge]].rect;
sqrt_z = complex_sqrt(z);
(*initial_tet)->corner[v[0]] = Infinity;
(*initial_tet)->corner[v[1]] = Zero;
(*initial_tet)->corner[v[2]] = complex_div(One, sqrt_z);
(*initial_tet)->corner[v[3]] = sqrt_z;
}
}
}